Linear Impulse – Momentum Relationship
Ft = mv = m(v2-v1)
Impulse (Ns)
Product of a force applied over a period of time
(Ft)
Momentum (kg m/s)
Quantity of motion. Product of mass * velocity
(mv)
Positive (negative) changes in Linear
Momentum are created by Net positive
(negative) Linear Impulse.
Course Reader:
Kinetics, p 48 - 53; Linear Impulse 53-61
LINEAR IMPULSE
Why?
• Mechanism for controlling linear velocity of the total
body center of mass
• Necessary for successful completion of general
locomotion tasks, and athletic movements
Vv1
Vv2
Vh1
Vh2
t
Ft = mv
= m(v2-v1)
= mv2 - mv1
Net Linear Impulse (F*t) Generation
BW
Fh
Fv
800
Ground Reaction Force (N)
Free Body
Diagram
Horizontal
600
Vertical
400
Positive Impulse
200
0
-200
Negative Impulse
-400
-600
-800
0.000
Net Vertical Force = Fv(+)+BW(-)
0.040
0.080
0.120
0.160
0.200
0.240
Time (s)
touchdown
take-off
Linear impulse magnitude = area under the
force-time curve, is dependent upon …
1) Ground reaction force magnitude (F)
2) ground contact duration (t)
Net Linear Impulse, the sum of negative and positive linear
impulse generated during the entire ground contact
phase (touchdown – take-off)
Fh
Fv
Ground Reaction Force (N)
BW
800
Ground reaction force (N)
Free Body
Diagram
Horizontal
600
Vertical
400
Positive Impulse
200
0
V1
-200
V2
Negative Impulse
-400
-600
-800
0.000
0.040
time=0
touchdown
0.080
Ft = mv
= m(v2-v1)
= mv2 - mv1
0.120
Time (s)
0.160
0.200
force=0
take-off
time (s)
0.240
How do you generate large Horizontal Impulse (force*time)?
– force, time, or a combination of force & time
• The mechanical goal of the task influences how Impulse is
generated
e.g. sprinters need to generate horizontal impulse quickly
Horizontal GRF (N)
1000
750
500
250
0
-250
-500
0.000
Time (s) after ground contact
0.100
0.200
0.300
Similar net changes in linear momentum can be
achieved with different force-time
linear impulse characteristics
Vh = 1.30 m/s
Vh = 1.29 m/s
Horizontal GRF (N)
1250
1000
750
500
250
0
-250
-500
0.000
0.050
0.100
0.150
0.200
time (s) after contact
0.250
Impulse-Momentum Relationship
Ft = HI = m(V2-V1)
Take-Off
Touchdown
mVh1
mVh2
1800
1600
1400
1200
1000
800
600
400
200
0
-200
-400
0.000
H GRF
V GRF
Fht
0.050
0.100
0.150
0.200
Time (s) after contact
0.250
Impulse-Momentum Relationship
Ft = HI = m(V2-V1)
Take-Off
Touchdown
mVv1
mVv2
1800
1600
1400
1200
1000
800
600
400
200
0
-200
-400
0.000
H GRF
V GRF
Fvt
0.050
0.100
0.150
0.200
Time (s) after contact
0.250
Calculating Net Linear Impulse Using Geometry
Take-Off
Touchdown
mVv1
mVh1
mVv2
mVh2
1800
1600
1400
1200
1000
800
600
400
200
0
-200
-400
0.000
H GRF
V GRF
0.050
0.100
0.150
Time (s) after contact
0.200
0.250
Back Somersault: Take-off Phase
Vv
Vh
Backwards
Rotation
Push
Tip
Load
Plate Departure
Needs:
Vertical Impulse (net positive),
Horizontal Impulse (net negative),
Backward-directed Angular Impulse
How?
Generation of Linear Impulse
During a Back Dive
Initiation
Take-Off
Near Zero Initial
TBCM
Momentum (mv)
Net Positive Vert. mv
Net Negative Horiz. mv
BACK Somersault
FH
FH
FV
time prior to take-off
FV
FR
take-off
Generation of Linear Impulse
During a Back Dive
Horizontal RF
2500
Vertical RF
2000
1500
Force (N)
BACK Somersault
VRF
1000
500
0
-0.5
FH
FH
FV
time prior to take-off
FV
FR
take-off
-0.4
-0.3
-0.2
-0.1
-500
0
Time Prior to Take-off (s)
time prior to take-off
take-off
Mechanical objective of the shot put:
• Vertical Impulse (net positive)
• Horizontal impulse (net negative - translate backward)
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F=ma
linear acceleration of the athlete’s center of mass is determined
by the sum of forces acting on the center of mass
Linear Impulse – Momentum Relationship
Ft = mv = m(v2-v1)
Free Body
Diagram
Vertical
Fv = FBW + Fv
Fv = mav
Fv = m (v/t)
Fv t = m (v)
(-)
FBW
Fh
Fv
Mass-Acceleration
Diagram
(+)
av
ah
F=ma
linear acceleration of the athlete’s center of mass is determined by the
sum of forces acting on the center of mass
Linear Impulse – Momentum Relationship
Ft = mv = m(v2-v1)
Free Body
Diagram
FBW
Fh
Fv
Horizontal
Fh = Fh(+)
Fh = mah
Fv = m (v/t)
Fv t = m (v)
Mass-Acceleration
Diagram
av
ah
Linear Impulse – Momentum Relationship
Ft = mv = m(v2-v1)
V GRF = BW
Vertical
force
Horizontal
force
BW
HGRF
VGRF
V GRF > BW
BW
V GRF = 0
BW
Vertical
force
Horizontal
force
BW
BW
BW
HGRF
VGRF
Body
weight
-0.500
-0.400
-0.300
-0.200
-0.100
Time (s) prior to departure
1200
1000
800
600
400
200
0
-200
-400
-600
0.000
Ground
Reaction
Forces
(Newtons)
Net Impulse = Change in Momentum
( Force) *(time) = (mass)*(velocity)
(+) vertical
impulse
Body weight
(-) horizontal
impulse
-0.500
-0.400
-0.300
-0.200
-0.100
Time (s) prior to departure
1200
1000
800
600
400
200
0
-200
-400
-600
0.000
Increase in the
positive vertical
velocity
Ground
Reaction Forces
(Newtons)
Increase in the
negative horizontal
velocity
Impulse
Momentum
Transfer
Projectile
motion
Mechanics of each phase influence the mechanics during
the next phase.
Impulse generation during the unseating phase will
influence initial conditions of the blocking phase.
Mechanical Objective of the Shot Put
Maximize the horizontal distance traveled by the shot
Projectile Motion
How does the shot become a projectile?
Total body momentum is generated and passed
on to the shot
Take-Home Message
Each foot (ground) contact is an
opportunity to:
a) increase,
b) decrease, or
c) maintain
your total body momentum.